Use L'Hospital to determine the following limit. Use exact values. \[ \lim _{x \rightarrow 0^{+}} x \ln x= \]

Step 1 ：The given limit is of the form 0*infinity, which is an indeterminate form. We can rewrite the expression as a fraction and then apply L'Hospital's rule. L'Hospital's rule states that the limit of a ratio of two functions as x approaches a certain value is equal to the limit of the ratios of their derivatives, provided the limit of the original functions is an indeterminate form.

Step 2 ：Rewrite the function as a ratio first: \(\frac{\ln x}{\frac{1}{x}}\).

Step 3 ：Find the derivative of the numerator and the denominator. The derivative of \(\ln x\) is \(\frac{1}{x}\) and the derivative of \(\frac{1}{x}\) is \(-\frac{1}{x^2}\).

Step 4 ：Apply L'Hospital's rule: \(\lim _{x \rightarrow 0^{+}} \frac{\ln x}{\frac{1}{x}} = \lim _{x \rightarrow 0^{+}} \frac{\frac{1}{x}}{-\frac{1}{x^2}}\).

Step 5 ：Simplify the expression to find the limit: \(\lim _{x \rightarrow 0^{+}} -x = 0\).

Step 6 ：The limit of the function as x approaches 0 from the right is \(\boxed{0}\).